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kolezko [41]
1 year ago
5

Describe what you would expect to observe, if anything, when a piece of nickel is placed in a solution of copper(II) nitrate. (s

elect all that apply)
a. Solid copper would form on the surface of the nickel metal.
b. The blue color of the solution would become darker.
c. Nickel atoms are converted to nickel ions in solution.
d. The blue color of the solution would become lighter.
e. No reaction would occur.
Chemistry
1 answer:
Tom [10]1 year ago
6 0

When a piece of nickel is placed in a solution of copper(II) nitrate:

a. Solid copper would form on the surface of the nickel metal.

c. Nickel atoms are converted to nickel ions in solution.

d. The blue color of the solution would become lighter.

Let's consider what happens when a piece of nickel is placed in a solution of copper(II) nitrate. Cu has a higher reduction potential than Ni, thus Cu will reduce and Ni will oxidize. This will be a single displacement redox reaction. The corresponding balanced molecular equation is:

Ni(s) + Cu(NO₃)₂(aq) ⇒ Ni(NO₃)₂(aq) + Cu(s)

The net ionic equation is:

Ni(s) + Cu²⁺(aq) ⇒ Ni²⁺(aq) + Cu(s)

<em>Describe what you would expect to observe, if anything, when a piece of nickel is placed in a solution of copper(II) nitrate.</em>

<em>a. Solid copper would form on the surface of the nickel metal.</em> YES. Solid copper will deposit on the surface of nickel metal.

<em>b. The blue color of the solution would become darker.</em> NO. The blue color, caused by Cu²⁺ ions, will become lighter, as their concentration decreases.

<em>c. Nickel atoms are converted to nickel ions in solution.</em> YES. Ni is oxidized to Ni²⁺.

<em>d. The blue color of the solution would become lighter.</em> YES. Due to the decrease in the concentration of Cu²⁺ ions.

<em>e. No reaction would occur.</em> NO.

When a piece of nickel is placed in a solution of copper(II) nitrate:

a. Solid copper would form on the surface of the nickel metal.

c. Nickel atoms are converted to nickel ions in solution.

d. The blue color of the solution would become lighter.

You can learn more about redox reactions here: brainly.com/question/13978139

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Write the net ionic equation to show the formation of a solid (insoluble salt) when the following solutions are mixed. Write nor
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<u>Answer:</u> The net ionic equations are written below.

<u>Explanation:</u>

Net ionic equation of any reaction does not include any spectator ions.

Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.

  • <u>For 1:</u>

The chemical equation for the reaction of calcium nitrate and sodium sulfate is given as:

Ca(NO_3)_2(aq.)+Na_2SO_4(aq.)\rightarrow CaSO_4(s)+2NaNO_3(aq.)

Ionic form of the above equation follows:

Ca^{2+}(aq.)+2NO_3^-(aq.)+2Na^+(aq.)+SO_4^{2-}(aq.)\rightarrow CaSO_4(s)+2Na^+(aq.)+2NO_3^-(aq.)

As, sodium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

Ca^{2+}(aq.)+SO_4^{2-}(aq.)\rightarrow CaSO_4(s)

  • <u>For 2:</u>

The chemical equation for the reaction of lead (II) nitrate and potassium chloride is given as:

Pb(NO_3)_2(aq.)+2KCl(aq.)\rightarrow PbCl_2(s)+2KNO_3(aq.)

Ionic form of the above equation follows:

Pb^{2+}(aq.)+2NO_3^-(aq.)+2K^+(aq.)+2Cl^{-}(aq.)\rightarrow PbCl_2(s)+2K^+(aq.)+2NO_3^-(aq.)

As, potassium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

Pb^{2+}(aq.)+2Cl^{-}(aq.)\rightarrow PbCl_2(s)

  • <u>For 3:</u>

The chemical equation for the reaction of calcium chloride and sodium phosphate is given as:

3CaCl_2(aq.)+2(NH_4)_3PO_4(aq.)\rightarrow Ca_3(PO_4)_2(s)+6NH_4Cl(aq.)

Ionic form of the above equation follows:

3Ca^{2+}(aq.)+6Cl^-(aq.)+6NH_4^+(aq.)+2PO_4^{3-}(aq.)\rightarrow Ca_3(PO_4)_2(s)+6NH_4^+(aq.)+6Cl^-(aq.)

As, ammonium and chloride ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

3Ca^{2+}(aq.)+2PO_4^{3-}(aq.)\rightarrow Ca_3(PO_4)_2(s)

  • <u>For 4:</u>

The chemical equation for the reaction of barium chloride and sodium sulfate is given as:

BaCl_2(aq.)+Na_2SO_4(aq.)\rightarrow BaSO_4(s)+2NaCl(aq.)

Ionic form of the above equation follows:

Ba^{2+}(aq.)+2Cl^-(aq.)+2Na^+(aq.)+SO_4^{2-}(aq.)\rightarrow BaSO_4(s)+2Na^+(aq.)+2Cl^-(aq.)

As, sodium and chloride ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

Ba^{2+}(aq.)+SO_4^{2-}(aq.)\rightarrow BaSO_4(s)

Hence, the net ionic equations are written above.

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